Let be $$T= \sum_{n=0}^N \hat{T} (n)e^{inx} $$

a trigonometric polynomial of grade $N$ without negative frequencies.

I wanna show that $$ \partial_x T= -iN(F_N \ast T-T) $$ Where $F_N \ast T$ meas the convolution of the Fejer Kernel and T.

might be easy..but I just can't work out the right conversion for this property..


$ \partial_x T= \sum_{n=0}^N \hat{T} (n) in e^{inx} $

$ =\sum_{n=0}^N ( \frac{1}{2 \pi} \int_{ -\pi}^{\pi} T(y)e^{-iny} dy ) e^{inx} in$

$ =\frac{1}{2 \pi} \int_{ -\pi}^{\pi} T(y) (\sum_{n=0}^N e^{in(x-y)})in $

from there on I get carried away in the wrong direction. is the derivative right?

Also..the Fejer Kernel can be expressed as the mean arithmetic value of the dirichtlet kernel so : $$ F_N= \frac{1}{n+1} \sum_{k=0}^n D_k(x) $$ Where $D_k = \sum_{n= -k}^k e^{inx} $ is the Dirichtlet Kernel

Very thankful for any help !


1). The last equality you wrote for the derivative is weird, since you have the $in$ outside a term that involves summing in $n$, so that's a mistake.

2). The definition of $F_N$ should involved '$N$' rather than '$n$'.

3). Either $F_N$ should be defined to be $\frac{1}{N}\sum_{k=0}^{N-1} D_k$ or the result you wish to prove should be $\partial_x T = -i(N+1)(F_N*T-T)$.

I will keep your definition of $F_N$ (with point (2) taken into account of course) and prove $\partial_x T = -i(N+1)(F_N*T-T)$. We show the fourier transform of both sides is the same -- this suffices.

Note $[-i(N+1)(F_N*T-T)]\hat{}(m) = -i(N+1)[\widehat{F_N}(m)\widehat{T}(m)-\widehat{T}(m)]$ and $\widehat{\partial_x T}(m) = im\widehat{T}(m)$ (the latter you can see from your first equality for the derivative). So, immediately, if $m < 0$ or if $m \ge N+1$, both sides are $0$ and we're good. So suppose $0 \le m \le N$. Then, $\widehat{F_N}(m) = \frac{1}{N+1}\sum_{k=0}^{N+1}\widehat{D_k}(m) = \frac{1}{N+1}\sum_{k=0}^{N+1} 1_{m \le k} = \frac{1}{N+1}[N+1-m]$, so we get $[-i(N+1)(F_N*T-T)]\hat{}(m) = -i(N+1)\widehat{T}(m)\frac{-m}{N+1} = im\widehat{T}(m)$, as desired.

  • $\begingroup$ thank you.. that did help a lot..didnt think of the idea of fourier transform the both sides... anyways I still have one qestion.. if i take $ \hat{F_N} (n) = \frac{1}{N} \sum_{k=0}^{N-1} \hat{D_k} (n) = \frac{1}{N} \sum_{k=0}^{N-1} \frac{1}{2 \pi} \int_{- \pi}^{ \pi} \sum_{m=-k}^{k} e^{im x} e^{in x} dx$ because of the orthogonality of the last part the integral equals 1. But then If i write out the sum it is $ \frac{1}{N} [N-1- |m| ] $ and the equasion doesnt hold anymore.. $\endgroup$ – wondering1123 Nov 19 '19 at 11:47
  • $\begingroup$ to sum up my question..what is $ \frac{1}{N} \sum_{k=0}^{N-1} 1$ als long as$ |m| \leq k $ ? $\endgroup$ – wondering1123 Nov 19 '19 at 12:43
  • $\begingroup$ @wondering1123 I'm sorry but I don't understand either of your comments. Your second comment has a typo but even if you meant "as long", I don't understand it. $\endgroup$ – mathworker21 Nov 19 '19 at 13:09
  • $\begingroup$ I don't understand the last bit of the second line in your proof where you calculate the fourier coefficient of the fejer kernel :-) $\endgroup$ – wondering1123 Nov 19 '19 at 13:54
  • $\begingroup$ Ok. I'm then confused by your first comment, since you're working with $\frac{1}{N}\sum_{k=0}^{N-1}$ when I very explicitly said (and wrote in my answer) that I'm working with $\frac{1}{N+1}\sum_{k=0}^{N+1}$. Anyways, it seems like you're just asking why $\sum_{k=0}^{N+1} 1_{|m| \le k} = N+1-|m|$ holds. Just to make sure we're on the same page, $1_{|m| \le k}$ denotes $1$ if $|m| \le k$ and $0$ otherwise. The numbers $k$ between $0$ and $N+1$ that satisfy $|m| \le k$ are $|m|,|m|+1,\dots,N+1$, provided $|m| \le N+1$ (which I'll add now as a separate case). $\endgroup$ – mathworker21 Nov 19 '19 at 13:57

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